Evaluating $\sum_{k=1}^{n} 2^{k}k$

Question

How can I evaulate the following series?

$$\sum_{k=1}^{n} 2^{k}k$$

I don't know where to begin. $2^k$ alone would be straight forward.

https://en.m.wikipedia.org/wiki/Arithmetico%E2%80%93geometric_sequence#Sum_of_the_terms — lab bhattacharjee, Jun 20 '19 at 15:46
Hint: Consider $S(x)=\sum\limits_{k=1}^n x^k$ and derivate with respect to $x$ once. What do you get?//(You are looking for $xS'(2))$. — Zacky, Jun 20 '19 at 15:47

score 3 · Accepted Answer · answered Jun 20 '19 at 15:46

3

Notice that $\sum_{k=1}^nk2^k=\sum_{m=1}^n\sum_{k=m}^n2^k$. Can you conclude from here?

answered Jun 20 '19 at 15:46

Leo163

2,647

He says in his question that he knows how to do geometric sums. – DanielV Jun 20 '19 at 15:48
They said $2^k$ alone would be straight forward – Jürgen Sukumaran Jun 20 '19 at 15:48

score 1 · Answer 2 · answered Jun 20 '19 at 15:57

Call $a_n$ the expression you want to get. First, we have $$a_{n+1}=a_n+2^{n+1}(n+1)$$

Second, we have $2a_{n}=\sum_{k=1}^n2^{k+1}k$. So, $$2a_{n}+\sum_{k=1}^n2^{k+1}=\sum_{k=1}^n2^{k+1}(k+1)=a_{n+1}-2$$ So, $2a_n+2^{n+2}-4=a_{n+1}-2$. Hence $$2a_{n}+2^{n+2}-2=a_{n+1}$$

We have two linear equations with the variables $a_{n+1}$,$a_n$. Solve it.

score 0 · Answer 3 · answered Jun 20 '19 at 15:48

0

Hint: Prove by induction that $$\sum_{k=1}^{n}2^k\cdot k=2\left(1+2^n(n-1)\right)$$

answered Jun 20 '19 at 15:48

Dr. Sonnhard Graubner

95,283

score -1 · Answer 4 · answered Jun 20 '19 at 15:51

-1

$$ \sum_{k=1}^{n} 2^{k} = infinity$$

then your sum is greater so it to goes to infinity

The comparison test shows why.

https://www.khanacademy.org/math/ap-calculus-bc/bc-series-new/bc-10-6/v/comparison-test-convergence

answered Jun 20 '19 at 15:51

aquagremlin

218

Is a finite sum, not an infinite sum. – Julian Mejia Jun 20 '19 at 16:05

Evaluating $\sum_{k=1}^{n} 2^{k}k$

4 Answers4